Sometimes substitutions like this can get you into trouble :
U^2 = X^3 => U = +/- X^ (3/2)
This is the part that I don’t know how to explain. How did we get from g’(x)=du/dx to du=g’(x)dx? Is there a rigorous proof for this? Is there a simple explanation that could be used for someone that knows little or nothing about derivatives and differentials that doesn’t involve “handwaving”? DPRK has had some suggestions but I’m afraid my audience would just return a blank stare. I have all summer to work on this.
Are you familiar with infinitesimals? Willing to use them?
They provide a perfectly rigorous explanation for your questions, and backed the original intuition behind calculus for both Newton and Leibniz. However, the existence of the infinitesimals was not rigorously established until the 60s. This is why most calculus pedagogy uses the (ε, δ) limit form, with all the baggage that entails (such as the claim that dy/dx is not a fraction, even though Leibniz certainly thought it was).
You aren’t going to prove the existence of the infinitesimals in an introductory class–but you aren’t going to establish the real numbers either. So that doesn’t seem like a big deal to me.
I know that they are rather small. Actually they are really, really, really, really small. They may be even smaller than that.
Then you are halfway there. The only other thing to remember is that they are not quite zero.
As I said above: it’s true by definition. At least, that’s the approach taken by all the standard Calculus textbooks I can remember seeing.
You can either define “du” to mean “the derivative of u with respect to x times dx,” or you can define it to mean something else and then prove/explain why that something else is equivalent to g’(x)dx.
I think it may be easier to think of a concrete example where there are actual physical quantities associated with x and u. Let us suppose that I am looking at an accelerating car and I want to know how far its gone after a particular time. In this case x is time in seconds and u is distance in meters and it happens that u=g(x)=x^2. Suppose I want to know how fast the car is going at a particular instant in time x.
One way I can do this is to at a particular time, x, I advance the time just one little tiny tiny smidgen, dx, and see how much the distance changes, lets call this change in distance du.
Well the distance at time=x was g(x)=x^2, and the distance of time (x+dx) was g(x+dx). So du defined as the change in distance was g(x+dx)-g(x).
Now we can work this out and find that
du=g(x+dx)-g(x)=(x+dx)^2-x^2 = x+2xdx+dx^2-2x = 2x dx+dx^2
Now dx is really really small, so dx^2 is really really really small, so small that you might as well ignore it. So we compute that du=2x dx, and in particular at that instant in time the distance it travels per unit time is du/dx=2x, which by amazing coincidence is g’(x).
Now suppose I have a wonky car that is doing all sorts of wonky things. So that its position over time is.
u=g(x) = x^3-5/x+ sin(x) -(e^x). I start out the same and get as far as,
du=g(x+dx)-g(x), and then divide both side by dx. So we get du/dx=(g(x+dx)-g(x))/dx, now computing all those differences in a pain in the ass so I’m not going to do that right now. Instead I’m going to just keep things in terms of functions.
Now remember that I wanted dx to be really really small, in fact I am going to take the limit as dx goes to zero.
du/dx = limit (g(x+dx)-g(x))/dx)
now to make thigs possibly more familiar I have going to change notation and say h=dx
so du/dx=limit (g(x+h)-g(x)/h) this right side is the definition of g’(x). So we see that for dx really really small du/dx=g’(x). Note you can’t argue with this step since without agreeing that its true we can’t discuss g’(x) at all.
now I use what I know about differentiation to compute that
g’(x)=3x^2-5ln(x)+cos(x)-e^x so if I move time ahead a little tiny smidgen dx the distance will change by du=(3x^2-5ln(x)+cos(x)-e^x)dx
I liked your idea of “concrete example “
Not sure if the limit even exists as x—>0 and not sure if you can integrate it. Are you sure this is a concrete example ?
How does Google Books decide which pages to let some user see and which pages produce the “That page is not in this preview” message?
(I’ve edited the URL above, changing ‘google.co.th’ to ‘google.com.’)
I’ve added color codes to the following excerpt. The beginning of the paragraph is on page 197 and shown in Gray. (The URL links to page 198.) The 12th-century comment alluding to quantities like dy/dx is shown in Red.
As to the mysterious “Indian sentence” you saw: I’ll guess you saw “That page is not in this preview” in Thai. (Try the new URL above.) I don’t know why Google Books lets me preview that page but not you. (Over half of the book’s pages are blocked to me also, but not 197-198.)
Blame me for the Thai error message (via the ‘co.th’ URL). Sometimes I manually edit those URLs to be ‘.com’ but Google changes them back to ‘co.th’ in the browser’s locator bar (unless I stay logged in to Google). I guess I figured that since Google changes English to Thai automatically for me, it would do the vice versa for you. 
(But that’s not how Google works. For me it leaves the English ‘co.uk’ intact; it just changes ‘.com’.)
Something to consider:
To the real numbers ℝ you may adjoin an infinitesimal element, let’s call it dx, which satisfies (dx)² = 0. This is a function on a formal infinitesimal line segment, a fattened point if you will, just as functions at a single point are given by a real number. dx is a coordinate function, just like x is on the real line.
Differential 1-forms on the real line now take the form g(x) dx, where g(x) is an ordinary smooth function. g(x) dx is not a function on the real line, rather a function on the space of infinitesimal paths in ℝ, i.e., something of the form f(x) + g(x) dx, with the f(x) = 0 so you just have g(x) dx.
Now let f(x) be an ordinary smooth function we wish to differentiate. Define a new function h(x,ε) = f(x+ε) (on the infinitesimal path space, so ε² = 0).
The derivative of f will now be obtained via taking h(x,ε) - f(x) = f(x+ε) - f(x). This will be a linear function of ε, namely εf’(x).
The corresponding 1-form is
(df)(x) = f(x + dx) - f(x) = f’(x) dx.
In summary, nothing that has not been said already in this thread, including that df = f’(x) dx very easily and that 1-forms are functions on an appropriate space (nothing non-rigorous about it). It still remains to make the connection to antidifferentiation and to definite integrals, but the differential forms that appear should not be confusing anybody.